Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1040

\[\boxed{\mathbf{1040}\mathbf{.}}\]

\[1)\ y = x^{2} - 2x + 2,\ \ \ x = 1,\ \ \ \]

\[касательная\ в\ точке\ x = 0;\]

\[y^{'}(x) = \left( x^{2} \right)^{'} - (2x - 2)^{'} =\]

\[= 2x - 2;\]

\[y^{'}(0) = 2 \bullet 0 - 2 = - 2;\]

\[y(0) = 0^{2} - 2 \bullet 0 + 2 = 2;\]

\[y = 2 - 2 \bullet (x - 0) = 2 - 2x.\]

\[S = \frac{1^{3}}{3} - \frac{0^{3}}{3} = \frac{1}{3}.\]

\[Ответ:\ \ \frac{1}{3}.\]

\[2)\ y = \frac{4}{x},\ \ \ y = 0,\ \ \ x = 6,\ \ \ \]

\[касательная\ в\ точке\ x = 2;\]

\[y^{'}(x) = 4 \bullet \left( \frac{1}{x} \right)^{'} = - \frac{4}{x^{2}};\]

\[y^{'}(2) = - \frac{4}{2^{2}} = - \frac{4}{4} = - 1;\]

\[y(2) = \frac{4}{2} = 2;\]

\[y = 2 - 1 \bullet (x - 2) =\]

\[= 2 - x + 2 = 4 - x.\]

\[\textbf{а)}\ \frac{4}{x} > 0\]

\[x > 0.\]

\[\textbf{б)}\ 4 - x > 0\ \]

\[x < 4.\]

\[S = \int_{2}^{6}{\frac{4}{x}\text{\ dx}} - \int_{2}^{4}{(4 - x)\text{\ dx}} =\]

\[= \left. \ \left( 4 \bullet \ln x \right) \right|_{2}^{6} - \left. \ \left( 4x - \frac{x^{2}}{2} \right) \right|_{2}^{4} =\]

\[= 4\ln 3 - 2.\]

\[Ответ:\ \ 4\ln 3 - 2.\]